I am unsure of Sohrab's process and assumptions as he is moving through the proof in the above quote ... could someone confirm (or otherwise) my interpretations as follows ... there are essentially 4 questions ( Q1, Q2, Q3 and Q4 respectively ...) ... ...

Second issue ... ... when Sohrab writes: "Indeed, if ##u \lt v## ... etc etc ... " ... ... he could have said ##u \gt v## ... but he is just taking ##u \lt v## as an example ... and we are left to infer that ##u \gt v## works similarly ... in other words there is no reason that ##u## is taken as less than ##v## as against taking ##v \lt u## ... ... Is that right? (Q2)

The above theorem concerns the Supremum Property, the Archimedean Property and the Nested Intervals Theorem ... so to give readers the context and notation regarding the above post I am posting the basic information on these properties/theorems ...

I'm only going to address Q1 and Q2 in this post. Really, since we know the diameter of the nested sequences converges to 0, we also know such a unique ##u## exists. Probably for pedagogical purposes, he shows that there cannot be two elements in the infinite intersection. Because these two elements ##u, v## are arbitrary anyway, there is no need to show both cases ##u<v## AND ##v<u##.

With regard to the fact that ##v-u>2^{-n}##, your reasoning is sound. It is easy to show that ##2^n>n## for all natural numbers. First, ##2^0=1>0##. Second, ##2^n>n \implies 2^{n+1}>n+1##. Adding 1 to the antecedent, ##2^n+1>n+1##. Since ##2 \cdot 2^n > 2^n+1## is true whenever ##n>0## (verify this by subtracting ##2^n## and taking ##\log_2##), we know that ##2^{n+1}>n+1## as desired.

This reasoning is a bit tedious, though. It's pretty obvious that ##(1/2^n)_{n \in \mathbb{N}} \to 0##,and any such sequence must get smaller than any given positive number—this is the spirit of the Archimedean property: there are no arbitrarily small numbers. Therefore, I imagine he took for granted that there existed a ##v-u>1/2^n## for some n.